In FEM is symmetry technically a boundary condition?

For example using the FEM on a symmetrically loaded and supported beam. The FEM can be simplified by using symmetry.

However, is this symmetry just implemented as a boundary condition or is it technically one? Because it's not actually constraining the movement given that the result is the same with or without it.

So my own impression is it's not a boundary condition?


2 Answers 2


The symmetry itself is not a boundary condition. It is a property of your system which means that both the geometry and the load are symmetric with respect to an axis or a plane. It allows to reduce the computation to a downsized domain, which leads to considerable computational time saving.
I guess you are using a FEA software and manually reduced the mesh. You then have to tell the software that some of the boundaries are not the physical boundaries of the domain you study, but are part of the plane of symmetry. Under the hood, the software applies some boundary conditions that are consequences of symmetry.
For example, if you have a planar symmetry, no displacement on the axis perpendicular to this plane is allowed, whereas only the rotation around this normal axis is permitted. So to answer your question, in "FEA software" language, symmetry is a boundary condition for the reasons I just explained, but this is not very rigorous.
The fact that you get the same results with and without the symmetry BC may be due to various reasons, one of which may be chance, if your load does not "put into use" the degrees of freedom blocked by the symmetry.


Symmetry is used to reduce the size of the object and therefore the mesh or allow more mesh to represent that reduced object.

So the “cut plane” of symmetry is not a boundary where the material changes in the real object.


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